Parabola

Need to 'massage' this around to vertex form of a parabola

x - y^2 + 8y = 13 + 6x - 6y

x - 6x = y^2 -14y+13

-5x = y^2 -14y+13     complete the square for y

-5x = ( y-7)^2 - 49 + 13

-5x = (y-7)^2 - 36

x = - 1/5 (y-7)^2 + 36/5            vertex is   7  , 36/5

Find the vertex of the graph of the equation x - y^2 + 8y = 13 + 6x - 6y.

Hello Guest!

$x - y^2 + 8y = 13 + 6x - 6y\\ x-6x+6y+8y-y^2-13=0\\ -y^2+14y-13=5x\\ x=f(y)=-\frac{1}{5}y^2+\frac{14}{5}y-\frac{13}{5}$

$\frac{df(y)}{dy}=-0.4y+2.8=0\\ y_v=\frac{-2.8}{-0.4}\\ \color{blue}y_v=7\\ x_v=-\frac{1}{5}y^2+\frac{14}{5}y-\frac{13}{5}$

$x_v=-\frac{1}{5}\cdot 7^2+\frac{14}{5}\cdot 7-\frac{13}{5}\\ \color{blue}x_v=7.2$

$The\ vertex\ of\ the\ parabola\ is\ \color{blue}P\ [7.2,\ 7]$

  !

I might as well put my 2 cents worth in as well  

Find the vertex of the graph of the equation x - y^2 + 8y = 13 + 6x - 6y.

$x - y^2 + 8y = 13 + 6x - 6y\\ -y^2+6y + 8y = 13 + 6x -x\\ - y^2+14y = 5x+13\\ y^2-14y = -5x-13\\ y^2-14y+49 = -5x-13+49\\ (y-7)^2 = -5x+36\\ (y-7)^2 = -5(x+7.2)\\ (y-7)^2 = -4*\frac{5}{4}(x+7.2)\\ $

This is a sideways parabola opening in the negative x direction.

It has a vertex of (-7.2,7)

It has a focal length of  5/4

So the focal point is  

The length of the latus r****m is 5

LaTex

x - y^2 + 8y = 13 + 6x - 6y\\
 -y^2+6y + 8y = 13 + 6x -x\\
- y^2+14y = 5x+13\\
 y^2-14y = -5x-13\\
 y^2-14y+49 = -5x-13+49\\
(y-7)^2 = -5x+36\\
(y-7)^2 = -5(x+7.2)\\
(y-7)^2 = -4*\frac{5}{4}(x+7.2)\\